Problem: Let $S$ be a surface in 3D described by the equation $z = x^4 - y^3 + x^2y - x + 2$. What is the equation of the plane tangent to $S$ at $(-1, -1)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $z = 4 - 3(x - 1) - 2(y - 1)$ (Choice B) B $z = 4 - 3(x + 1) - 2(y + 1)$ (Choice C) C $z = 4 - (x - 7) - (y - 2)$ (Choice D) D $z = 4 + 13(x + 1) + 6(y + 1)$
Solution: The equation for a tangent plane of an explicitly defined surface $z = f(x, y)$ at the point $(a, b)$ is: $f(a, b) + f_x(x - a) + f_y(y - b) = z$ [What's the intuition behind the formula?] Let's find $f(-1, -1)$, $f_x$, and $f_y$. $\begin{aligned} &f(-1, -1) = 1 + 1 - 1 + 1 + 2 = 4 \\ \\ &f_x = 4x^3 + 2xy - 1 = -3 \\ \\ &f_y = -3y^2 + x^2 = -2 \end{aligned}$ Putting it all together, here's the equation for the tangent plane of $S$ at $(-1, -1)$ : $z = 4 - 3(x + 1) - 2(y + 1)$